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  1. K

    was Re:Ir-remote Now is: Op Amp discussion

    You have to investigate the circuit more closely. The opamp is rated for 20mA, which is what the circuit demands. Look at the ouput of an opamp, the emitter follower. You can easily approximate the current because of the diodes in parallel with the resistors. Notice that the load resistor is in...
  2. K

    MUX and DEMUX

    Why aren't multiplaxars and demultiplexars used for serial to parallel conversions? I understand that the operation of these devices could be used for conversions. Any help as to why would be appreciated. Are they used in chips to help complete these operations? I would like to see a...
  3. K

    Serial to Parallel Output

    Does the UART require programming? I thought it was just configured. What sort of programming is required?
  4. K

    was Re:Ir-remote Now is: Op Amp discussion

    The input signal as well as the output signal and the DC bias looks correct. What is wrong with 18ma load current? The gain is a little higher than RF over RI but the gain is much lower than open loop.
  5. K

    was Re:Ir-remote Now is: Op Amp discussion

    The 1Kohm to -18 can simulate a load. The current is even close enough. But more importantly, it shows that this is a collector resistor. Whenever you are able to reduce the voltage away from a rail, such as at 0volts, you are reducing the gain. Just apply the signal and notice the gain is much...
  6. K

    help: transceiver switch

    It sounds like you want to use a simple logic device but with an analog signal. Maybe an analog switch. Check to see if you can't use a digital device to enable the analog ouput. Otherwise, you could probably use a couple of emitter followers that are enabled or disabled with logic. You might...
  7. K

    was Re:Ir-remote Now is: Op Amp discussion

    Audioguru, we are not talking about rail to rail operation. We are in fact able to reduce the gain with the feedback resistor because the opamp is a symmetrical device. The positive feeback resistor can lower the gain because it is infact still a collector resistor. You will notice that it works...
  8. K

    OUTPUT IMPEDANCE

    This is where I have trouble with the datasheets. Sometimes the data can be irrelevant. In this case, they should show the output portion of the circuit with a signal applied. That way you could add your resistance and determine how it will affect the output. What they are actually saying is...
  9. K

    inverter oscillator

    What I find amazing is how the transition will allow for the charging of the capacitor. The transition is of a certain duration. If I use a long time constant, I think that the transition is of a longer duration. This is how I get an oscillation at the low frequency.
  10. K

    -2mV ~ 2mV signal

    You need to gain first, then filter. The filter should be simple construction, not to say that you couldn't use a capacitor around the opamp like a state variable filter. You can use many stages so that each stage will properly handle the amplitude of the signal which is continuing to be...
  11. K

    was Re:Ir-remote Now is: Op Amp discussion

    One might get a better feel for opamps if you connect the negative feedback from ouput to noninverting input instead of the inverting input. This is called positive feedback, but the idea of gain is relatively the same. In other words, connect the opamp backwards and realize the similarities.
  12. K

    RJ45 network card to IR

    Referring to the original circuit found on the home page. I can see that you are using the 5volts respective to 9volts through the infrared diode. I don't think this is a reliable situation. I could go into detail as to why, but I will leave it at that unless you are interested. The signal, by...
  13. K

    inverter oscillator

    Audioguru, thank you for describing my circuit. I used a pullup because the example in my book uses a pullup. The pullup is on the output of the first inverter. I did not happen to notice that I constructed a classic CMOS oscillator, though it appears to be. I have this clock hooked to a BCD...
  14. K

    inverter oscillator

    The pullup is on the output of the first inverter. By the way, I stumbled upon the fix. I used a 10M resistor from the input of the second inverter to it's output. I would have thought that a resistor to ground on the input of the second inverter would work. But it seems that the complimetary...
  15. K

    Design of DAC

    The idea of a D/A is surrounded by the weighted ladder. Each sequence is able to produce a current that results in a voltage. But I am wary of this opamp configuration because of how it is used. The opamp is meant to operate high gain, even though it will operate as a constant current source. I...
  16. K

    BCD counter

    I was looking for an ordinary counter, but I only found a BCD counter. It turns out that it works the way I wanted it to. I had planned on using a regular counter fed to a decoder. Each sequence would decode one output. This is what this device does. Maybe we should call this a counter/decoder...
  17. K

    Battery back-up

    That sounds like a simple switch, but I won't ask why you need to distinguish between the two supplies. You can activate and deactive a circuit using a semiconductor if the voltages allow. I would correct the voltage in order to use a simple semiconductor rather than use a relay.
  18. K

    inverter oscillator

    I built an inverter oscillator with two inverters connected by a capacitor and a pullup on the first inverter. The output of the second inverter is fed to the input of the first inverter. The problem I'm having is that it will only oscillate when I touch the side of the capacitor, which shortens...
  19. K

    pulse width modulation

    I will have to disagree with you about the thresholds. I don't think they are necessary in deciphering the operation of the circuit. But I will continue to read your posting to determine otherwise.
  20. K

    pulse width modulation

    I think I can corroborate. Let's start with what we can easily agree upon. The output of the second inverter goes through the capacitor. This is the first observation. The capacitor will charge to 5 volts, bringing the high that went through the capacitor to 0 volts. Are these two observations...
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